Question #0c86a

1 Answer
Jun 21, 2016

Discussed below

Explanation:

Given

#h->"The height from which the canon ball was fired"=60m#

#u->"The vlocity of projection"=25m/s#

#alpha -> "Angle of projection"=53^@#

#"And "sin53^@=0.8 and cos53^@=0.6#

Now

#"Horizontal component of vel.of projection " ucos53^@=25*0.6=15m/s#

#"Vertical component of vel.of projection " usin53^@=25*0.8=20m/s#

At the time when it reaches at maximum height H from its point of projection its vertical component becomes zero, So considering #g=ms^-2# we get

#0^2=20^2-2*10*H=>H=20m#

Hence at its maximum height the total height fron ground
#H_"total"=60+20=80m#

Now if its time of flight be t s
then it will undergo a vertical displacement downward during this time. So
-
#-60=20*t-1/2*20*t^2#

#=>t^2-2t-6=0#

#=>t=(2+sqrt(4-4*(-6)*1))/2#

#=>t=(sqrt7+1)s#

Net horizontal displacement during this time of flight

#=ucosalpha*t=15(sqrt7+1)m#